Solution

Let $x^{3}=t$

$\therefore 3 x^{2} d x=d t$

$ \begin{aligned} \Rightarrow \int \frac{3 x^{2}}{x^{6}+1} d x & =\int \frac{d t}{t^{2}+1} \\ & =\tan ^{1} t+C \\ & =\tan ^{-1}(x^{3})+C \end{aligned} $

Solution

Let $2 x=t$

$\therefore 2 d x=d t$ $\Rightarrow \int \frac{1}{\sqrt{1+4 x^{2}}} d x=\frac{1}{2} \int \frac{d t}{\sqrt{1+t^{2}}}$

$ \begin{matrix} =\frac{1}{2}[\log |t+\sqrt{t^{2}+1}|]+C & {[\int \frac{1}{\sqrt{x^{2}+a^{2}}} d t=\log |x+\sqrt{x^{2}+a^{2}}|]} \\ =\frac{1}{2} \log |2 x+\sqrt{4 x^{2}+1}|+C & \end{matrix} $

Solution

Let $2-x=t$

$\Rightarrow-d x=d t$

$ \begin{aligned} \Rightarrow \int \frac{1}{\sqrt{(2-x)^{2}+1}} d x & =-\int \frac{1}{\sqrt{t^{2}+1}} d t \\ & =-\log |t+\sqrt{t^{2}+1}|+C \quad[\int \frac{1}{\sqrt{x^{2}+a^{2}}} d t=\log |x+\sqrt{x^{2}+a^{2}}|] \\ & =-\log |2-x+\sqrt{(2-x)^{2}+1}|+C \\ & =\log |\frac{1}{(2-x)+\sqrt{x^{2}-4 x+5}}|+C \end{aligned} $

Solution

Let $5 x=t$ $\therefore 5 d x=d t$

$ \begin{aligned} \Rightarrow \int \frac{1}{\sqrt{9-25 x^{2}}} d x & =\frac{1}{5} \int \frac{1}{9-t^{2}} d t \\ & =\frac{1}{5} \int \frac{1}{\sqrt{3^{2}-t^{2}}} d t \\ & =\frac{1}{5} \sin ^{-1}(\frac{t}{3})+C \\ & =\frac{1}{5} \sin ^{-1}(\frac{5 x}{3})+C \end{aligned} $

Solution

Let $\sqrt{2} x^{2}=t$

$\therefore 2 \sqrt{2} x d x=d t$

$\Rightarrow \int \frac{3 x}{1+2 x^{4}} d x=\frac{3}{2 \sqrt{2}} \int \frac{d t}{1+t^{2}}$

$ =\frac{3}{2 \sqrt{2}}[\tan ^{-1} t]+C $

$=\frac{3}{2 \sqrt{2}} \tan ^{-1}(\sqrt{2} x^{2})+C$

Solution

Let $x^{3}=t$ $\therefore 3 x^{2} d x=d t$

$$ \begin{aligned} \Rightarrow \int \frac{x^{2}}{1-x^{6}} d x & =\frac{1}{3} \int \frac{d t}{1-t^{2}} \\ & =\frac{1}{3}[\frac{1}{2} \log |\frac{1+t}{1-t}|]+C \\ & =\frac{1}{6} \log |\frac{1+x^{3}}{1-x^{3}}|+C \end{aligned} $$

Solution

$$ \begin{equation*} \int \frac{x-1}{\sqrt{x^{2}-1}} d x=\int \frac{x}{\sqrt{x^{2}-1}} d x-\int \frac{1}{\sqrt{x^{2}-1}} d x \tag{1} \end{equation*} $$

For $\int \frac{x}{\sqrt{x^{2}-1}} d x$, let $x^{2}-1=t \Rightarrow 2 x d x=d t$

$\therefore \int \frac{x}{\sqrt{x^{2}-1}} d x=\frac{1}{2} \int \frac{d t}{\sqrt{t}}$

$=\frac{1}{2} \int t^{-\frac{1}{2}} d t$

$=\frac{1}{2}[2 t^{\frac{1}{2}}]$

$=\sqrt{t}$

$=\sqrt{x^{2}-1}$

From (1), we obtain

$ \begin{aligned} \int \frac{x-1}{\sqrt{x^{2}-1}} d x & =\int \frac{x}{\sqrt{x^{2}-1}} d x-\int \frac{1}{\sqrt{x^{2}-1}} d x \quad \quad[\int \frac{1}{\sqrt{x^{2}-a^{2}}} d t=\log |x+\sqrt{x^{2}-a^{2}}|] \\ & =\sqrt{x^{2}-1}-\log |x+\sqrt{x^{2}-1}|+C \end{aligned} $

Solution

Let $x^{3}=t$

$\Rightarrow 3 x^{2} d x=d t$

$ \begin{aligned} \therefore \int \frac{x^{2}}{\sqrt{x^{6}+a^{6}}} d x & =\frac{1}{3} \int \frac{d t}{\sqrt{t^{2}+(a^{3})^{2}}} \\ & =\frac{1}{3} \log |t+\sqrt{t^{2}+a^{6}}|+C \\ & =\frac{1}{3} \log |x^{3}+\sqrt{x^{6}+a^{6}}|+C \end{aligned} $

Solution

Let $\tan x=t$

$\therefore \sec ^{2} x d x=d t$

$ \begin{aligned} \Rightarrow \int \frac{\sec ^{2} x}{\sqrt{\tan ^{2} x+4}} d x & =\int \frac{d t}{\sqrt{t^{2}+2^{2}}} \\ & =\log |t+\sqrt{t^{2}+4}|+C \\ & =\log |\tan x+\sqrt{\tan ^{2} x+4}|+C \end{aligned} $

Solution

$ \int \frac{1}{\sqrt{x^{2}+2 x+2}} d x=\int \frac{1}{\sqrt{(x+1)^{2}+(1)^{2}}} d x $

Let $x+1=t$

$\therefore d x=d t$

$\Rightarrow \int \frac{1}{\sqrt{x^{2}+2 x+2}} d x=\int \frac{1}{\sqrt{t^{2}+1}} d t$

$ \begin{aligned} & =\log |t+\sqrt{t^{2}+1}|+C \\ & =\log |(x+1)+\sqrt{(x+1)^{2}+1}|+C \\ & =\log |(x+1)+\sqrt{x^{2}+2 x+2}|+C \end{aligned} $

Solution

$ \int \frac{1}{9 x^{2}+6 x+5} d x=\int \frac{1}{(3 x+1)^{2}+(2)^{2}} d x $

Let $(3 x+1)=t$

$\therefore 3 d x=d t$

$\Rightarrow \int \frac{1}{(3 x+1)^{2}+(2)^{2}} d x=\frac{1}{3} \int \frac{1}{t^{2}+2^{2}} d t$

$ \begin{aligned} & =\frac{1}{3}[\frac{1}{2} \tan ^{-1}(\frac{t}{2})]+C \\ & =\frac{1}{6} \tan ^{-1}(\frac{3 x+1}{2})+C \end{aligned} $

Solution

$$ \begin{aligned} & =\int \frac{1}{\sqrt{7-(x+3)^2-(3)^2}} d x \\ & =\int \frac{1}{\sqrt{7+9-(x+3)^2}} d x \\ & =\int \frac{1}{\sqrt{16-(x+3)^2}} d x \\ & =\int \frac{1}{\sqrt{(4)^2-(x+3)^2}} d x \end{aligned} $$

It is of form $$ \int \frac{d x}{\sqrt{a^2-x^2}}=\sin ^{-1} \frac{x}{a}+C $$ $\therefore$ Replacing $x$ by $a$ by 4 , we get $$ =\sin ^{-1}\left(\frac{x+3}{4}\right)+C $$

Solution

$(x-1)(x-2)$ can be written as $x^{2}-3 x+2$.

Therefore,

$x^{2}-3 x+2$

$=x^{2}-3 x+\frac{9}{4}-\frac{9}{4}+2$

$=(x-\frac{3}{2})^{2}-\frac{1}{4}$

$=(x-\frac{3}{2})^{2}-(\frac{1}{2})^{2}$

$\therefore \int \frac{1}{\sqrt{(x-1)(x-2)}} d x=\int \frac{1}{\sqrt{(x-\frac{3}{2})^{2}-(\frac{1}{2})^{2}}} d x$

Let $x-\frac{3}{2}=t$

$\therefore d x=d t$

$\Rightarrow \int \frac{1}{\sqrt{(x-\frac{3}{2})^{2}-(\frac{1}{2})^{2}}} d x=\int \frac{1}{\sqrt{t^{2}-(\frac{1}{2})^{2}}} d t$

$=\log |t+\sqrt{t^{2}-(\frac{1}{2})^{2}}|+C$

$=\log |(x-\frac{3}{2})+\sqrt{x^{2}-3 x+2}|+C$

Solution

$8+3 x-x^{2}$ can be written as $8-(x^{2}-3 x+\frac{9}{4}-\frac{9}{4})$.

Therefore,

$8-(x^{2}-3 x+\frac{9}{4}-\frac{9}{4})$

$=\frac{41}{4}-(x-\frac{3}{2})^{2}$

$\Rightarrow \int \frac{1}{\sqrt{8+3 x-x^{2}}} d x=\int \frac{1}{\sqrt{\frac{41}{4}-(x-\frac{3}{2})^{2}}} d x$

Let $x-\frac{3}{2}=t$

$\therefore d x=d t$

$\Rightarrow \int \frac{1}{\sqrt{\frac{41}{4}-(x-\frac{3}{2})^{2}}} d x=\int \frac{1}{\sqrt{(\frac{\sqrt{41}}{2})^{2}-t^{2}}} d t$

$=\sin ^{-1}(\frac{t}{\frac{\sqrt{41}}{2}})+C$

$=\sin ^{-1}(\frac{x-\frac{3}{2}}{\frac{\sqrt{41}}{2}})+C$

$=\sin ^{-1}(\frac{2 x-3}{\sqrt{41}})+C$

Solution

$$ \begin{aligned} & =\int \frac{1}{\sqrt{x^2-2(x)\left(\frac{a+b}{2}\right)+\left(\frac{a+b}{2}\right)^2-\left(\frac{a+b}{2}\right)^2+a b}} d x \\ & =\int \frac{1}{\sqrt{\left(x-\frac{a+b}{2}\right)^2-\left(\frac{a+b}{2}\right)^2+a b}} d x \\ & =\int \frac{1}{\sqrt{\left(x-\frac{a+b}{2}\right)^2-\left(\frac{a^2+b^2+2 a b}{4}\right)+a b}} d x \\ & =\int \frac{1}{\sqrt{\left(x-\frac{a+b}{2}\right)^2+\frac{-a^2-b^2-2 a b+4 a b}{4}}} d x \\ & =\int \frac{1}{\sqrt{\left(x-\frac{a+b}{2}\right)^2+\frac{-a^2-b^2+2 a b}{4}}} d x \end{aligned} $$

Solution

Let $4 x+1=A \frac{d}{d x}(2 x^{2}+x-3)+B$

$\Rightarrow 4 x+1=A(4 x+1)+B$

$\Rightarrow 4 x+1=4 A x+A+B$

Equating the coefficients of $x$ and constant term on both sides, we obtain $4 A=4 \Rightarrow A=1$

$A+B=1 \Rightarrow B=0$

Let $2 x^{2}+x-3=t$

$\therefore(4 x+1) d x=d t$

$ \begin{aligned} \Rightarrow \int \frac{4 x+1}{\sqrt{2 x^{2}+x-3}} d x & =\int \frac{1}{\sqrt{t}} d t \\ & =2 \sqrt{t}+C \\ & =2 \sqrt{2 x^{2}+x-3}+C \end{aligned} $

Solution

Let $x+2=A \frac{d}{d x}(x^{2}-1)+B$

$\Rightarrow x+2=A(2 x)+B$

Equating the coefficients of $x$ and constant term on both sides, we obtain

$2 A=1 \Rightarrow A=\frac{1}{2}$

$B=2$

From (1), we obtain $(x+2)=\frac{1}{2}(2 x)+2$

Then, $\int \frac{x+2}{\sqrt{x^{2}-1}} d x=\int \frac{\frac{1}{2}(2 x)+2}{\sqrt{x^{2}-1}} d x$

$$ \begin{equation*} =\frac{1}{2} \int \frac{2 x}{\sqrt{x^{2}-1}} d x+\int \frac{2}{\sqrt{x^{2}-1}} d x \tag{2} \end{equation*} $$

In $\frac{1}{2} \int \frac{2 x}{\sqrt{x^{2}-1}} d x$, let $x^{2}-1=t \Rightarrow 2 x d x=d t$

$ \begin{aligned} \frac{1}{2} \int \frac{2 x}{\sqrt{x^{2}-1}} d x & =\frac{1}{2} \int \frac{d t}{\sqrt{t}} \\ & =\frac{1}{2}[2 \sqrt{t}] \\ & =\sqrt{t} \\ & =\sqrt{x^{2}-1} \end{aligned} $

Then, $\int \frac{2}{\sqrt{x^{2}-1}} d x=2 \int \frac{1}{\sqrt{x^{2}-1}} d x=2 \log |x+\sqrt{x^{2}-1}|$

From equation (2), we obtain

$ \int \frac{x+2}{\sqrt{x^{2}-1}} d x=\sqrt{x^{2}-1}+2 \log |x+\sqrt{x^{2}-1}|+C $

$$ \begin{aligned} & \int \frac{5 x-2}{1+2 x+3 x^2} d x \\ & =5 \int \frac{x-\frac{2}{5}}{1+2 x+3 x^2} d x \\ & =\frac{5}{6} \int \frac{6 x-\frac{12}{5}}{1+2 x+3 x^2} d x \\ & =\frac{5}{6} \int \frac{6 x+2-\frac{12}{5}-2}{1+2 x+3 x^2} d x \\ & =\frac{5}{6} \int \frac{6 x+2-\frac{22}{5}}{1+2 x+3 x^2} d x \end{aligned} $$

$$ \begin{gathered} =\frac{5}{6} \int \frac{6 x+2}{1+2 x+3 x^2} d x-\frac{5}{6} \times \frac{22}{5} \int \frac{d x}{1+2 x+3 x^2} \\ =\frac{5}{6} \int \frac{6 x+2}{1+2 x+3 x^2} d x-\frac{11}{3} \int \frac{d x}{1+2 x+3 x^2} \\ \downarrow \end{gathered} $$

Solving $I_1$ $$ \mathrm{I}_1=\frac{5}{6} \int \frac{2+6 x}{3 x^2+2 x+1} d x $$

Let $$ 3 x^2+2 x+1=t $$

Diff both sides w.r.t.x $$ 6 x+2+0=\frac{d t}{d x} $$

$$ d x=\frac{d t}{6 x+2} $$

Thus, our equation becomes $$ \mathrm{I}_1=\frac{5}{6} \int \frac{2+6 x}{3 x^2+2 x+1} d x $$

Put the values of $\left(3 x^2+2 x+1\right)$ and $d x$, we get

$$ \begin{aligned} & \mathrm{I}_1=\frac{5}{6} \int \frac{6 x+2}{t} d x \\ & \mathrm{I}_1=\frac{5}{6} \int \frac{6 x+2}{t} \times \frac{d t}{6 x+2} \\ & \mathrm{I}_1=\frac{5}{6} \int \frac{1}{t} \cdot d t \\ & \mathrm{I}_1=\frac{5}{6} \log |t|+C_1 \\ & \mathrm{I}_1=\frac{5}{6} \log \left|3 x^2+2 x+1\right|+C_1 \quad\left(\text { Using } t=3 x^2+2 x+1\right) \end{aligned} $$

$\left(U\right.$ sing $\left.t=3 x^2+2 x+1\right)$

$$ \begin{aligned} & =\frac{11}{9} \cdot \frac{3}{\sqrt{2}} \tan ^{-1}\left(\frac{3 x+1}{\sqrt{2}}\right)+\frac{11}{9} C_2 \\ & =\frac{11}{3 \sqrt{2}} \tan ^{-1}\left(\frac{3 x+1}{\sqrt{2}}\right)+C_3 \end{aligned} $$ (Where $C_3=\frac{11}{9} C_2$ )

Now, putting the values of $\mathrm{I}_1$ and $\mathrm{I}_2$ in (1) $$ \begin{aligned} & \int \frac{5 x-2}{1+2 x+3 x^2} \cdot d x \\ & \quad=\frac{5}{6} \int \frac{2+6 x}{1+2 x+3 x^2} \cdot d x-\frac{11}{3} \int \frac{1}{1+2 x+3 x^2} \cdot d x \\ & =\frac{5}{6} \log \left|3 x^2+2 x+1\right|-\frac{11}{3 \sqrt{2}} \tan ^{-1}\left(\frac{3 x+1}{\sqrt{2}}\right)+C \end{aligned} $$

Solution

$\frac{6 x+7}{\sqrt{(x-5)(x-4)}}=\frac{6 x+7}{\sqrt{x^{2}-9 x+20}}$

Let $6 x+7=A \frac{d}{d x}(x^{2}-9 x+20)+B$

$\Rightarrow 6 x+7=A(2 x-9)+B$

Equating the coefficients of $x$ and constant term, we obtain

$2 A=6 \Rightarrow A=3$

$-9 A+B=7 \Rightarrow B=34$

$\therefore 6 x+7=3(2 x-9)+34$

$ \begin{aligned} \int \frac{6 x+7}{\sqrt{x^{2}-9 x+20}} & =\int \frac{3(2 x-9)+34}{\sqrt{x^{2}-9 x+20}} d x \\ & =3 \int \frac{2 x-9}{\sqrt{x^{2}-9 x+20}} d x+34 \int \frac{1}{\sqrt{x^{2}-9 x+20}} d x \end{aligned} $

Let $I_1=\int \frac{2 x-9}{\sqrt{x^{2}-9 x+20}} d x$ and $I_2=\int \frac{1}{\sqrt{x^{2}-9 x+20}} d x$

$\therefore \int \frac{6 x+7}{\sqrt{x^{2}-9 x+20}}=3 I_1+34 I_2$

Then,

$I_1=\int \frac{2 x-9}{\sqrt{x^{2}-9 x+20}} d x$

Let $x^{2}-9 x+20=t$

$\Rightarrow(2 x-9) d x=d t$

$\Rightarrow I_1=\frac{d t}{\sqrt{t}}$

$I_1=2 \sqrt{t}$

$I_1=2 \sqrt{x^{2}-9 x+20}$

and $I_2=\int \frac{1}{\sqrt{x^{2}-9 x+20}} d x$ $x^{2}-9 x+20$ can be written as $x^{2}-9 x+20+\frac{81}{4}-\frac{81}{4}$.

Therefore,

$x^{2}-9 x+20+\frac{81}{4}-\frac{81}{4}$

$=(x-\frac{9}{2})^{2}-\frac{1}{4}$

$=(x-\frac{9}{2})^{2}-(\frac{1}{2})^{2}$

$\Rightarrow I_2=\int \frac{1}{\sqrt{(x-\frac{9}{2})^{2}-(\frac{1}{2})^{2}}} d x$

$I_2=\log |(x-\frac{9}{2})+\sqrt{x^{2}-9 x+20}|$

Substituting equations (2) and (3) in (1), we obtain

$ \begin{aligned} \int \frac{6 x+7}{\sqrt{x^{2}-9 x+20}} d x & =3[2 \sqrt{x^{2}-9 x+20}]+34 \log [(x-\frac{9}{2})+\sqrt{x^{2}-9 x+20}]+C \\ & =6 \sqrt{x^{2}-9 x+20}+34 \log [(x-\frac{9}{2})+\sqrt{x^{2}-9 x+20}]+C \end{aligned} $

Solution

Let $x+2=A \frac{d}{d x}(4 x-x^{2})+B$

$\Rightarrow x+2=A(4-2 x)+B$

Equating the coefficients of $x$ and constant term on both sides, we obtain $-2 A=1 \Rightarrow A=-\frac{1}{2}$

$4 A+B=2 \Rightarrow B=4$

$\Rightarrow(x+2)=-\frac{1}{2}(4-2 x)+4$

$\therefore \int \frac{x+2}{\sqrt{4 x-x^{2}}} d x=\int \frac{-\frac{1}{2}(4-2 x)+4}{\sqrt{4 x-x^{2}}} d x$ $=-\frac{1}{2} \int \frac{4-2 x}{\sqrt{4 x-x^{2}}} d x+4 \int \frac{1}{\sqrt{4 x-x^{2}}} d x$

Let $I_1=\int \frac{4-2 x}{\sqrt{4 x-x^{2}}} d x$ and $I_2 \int \frac{1}{\sqrt{4 x-x^{2}}} d x$

$\therefore \int \frac{x+2}{\sqrt{4 x-x^{2}}} d x=-\frac{1}{2} I_1+4 I_2$

Then, $I_1=\int \frac{4-2 x}{\sqrt{4 x-x^{2}}} d x$

Let $4 x-x^{2}=1$

$\Rightarrow(4-2 x) d x=d t$

$\Rightarrow l_1=\int _{\sqrt{t}}^{d t}=2 \sqrt{t}=2 \sqrt{4 x-x^{2}}$

$I_2=\int \frac{1}{\sqrt{4 x-x^{2}}} d x$

$\Rightarrow 4 x-x^{2}=-(-4 x+x^{2})$

$=(-4 x+x^{2}+4-4)$

$=4-(x-2)^{2}$

$=(2)^{2}-(x-2)^{2}$

$\therefore I_2=\int \frac{1}{\sqrt{(2)^{2}-(x-2)^{2}}} d x=\sin ^{-1}(\frac{x-2}{2})$

Using equations (2) and (3) in (1), we obtain

$ \begin{aligned} \int \frac{x+2}{\sqrt{4 x-x^{2}}} d x & =-\frac{1}{2}(2 \sqrt{4 x-x^{2}})+4 \sin ^{-1}(\frac{x-2}{2})+C \\ & =-\sqrt{4 x-x^{2}}+4 \sin ^{-1}(\frac{x-2}{2})+C \end{aligned} $

Solution

$$ I=\int \frac{x+2}{\sqrt{x^2+2 x+3}} d x $$

Let, $x+2=\lambda \frac{d}{d x}\left(x^2+2 x+3\right)+\mu$, then $$ x+2=(2 x+2) \lambda+\mu $$

Comparing coefficients of like powers of $x$, we get and $2 \lambda=1$ or $\lambda=1 / 2$ or $$ \begin{aligned} 2 \lambda+\mu =2 \ 1+\mu =2 \end{aligned} $$

or $\quad \mu=1$ Hence, $\quad I=\int \frac{x+2}{\sqrt{x^2+2 x+3}} d x$

$$ \begin{aligned} & \qquad=\int \frac{\frac{1}{2}(2 x+2)+1}{\sqrt{x^2+2 x+3}} d x \\ & =\frac{1}{2} \int \frac{(2 x+2)}{\sqrt{x^2+2 x+3}} d x+\int \frac{1}{\sqrt{x^2+2 x+3}} d x \\ & =\int \frac{d t}{2 \sqrt{t}}+\int \frac{d x}{\sqrt{\left(x^2\right)+2 x+1+2}} \quad \text { where } \quad t=x^2+2 x+3 \end{aligned} $$

$$ \begin{aligned} & \text { or } d t=(2 x+2) d x \\ & =\sqrt{t}+\int \frac{d x}{\sqrt{(x+1)^2+(\sqrt{2})^2}} \\ & =\sqrt{x^2+2 x+3}+\log \left\lvert, \frac{(x+1)}{2}+\sqrt{(x+1)^2+(\sqrt{2})^2}+C\right. \\ & \left.=\sqrt{x^2+2 x+3}+\log \frac{(x+1)}{2}+\sqrt{x^2+2 x+3} \right\rvert,+C \end{aligned} $$

Solution

Let $(x+3)=A \frac{d}{d x}(x^{2}-2 x-5)+B$

$(x+3)=A(2 x-2)+B$

Equating the coefficients of $x$ and constant term on both sides, we obtain

$ \begin{aligned} & 2 A=1 \Rightarrow A=\frac{1}{2} \\ & -2 A+B=3 \Rightarrow B=4 \\ & \therefore(x+3)=\frac{1}{2}(2 x-2)+4 \\ & \Rightarrow \int \frac{x+3}{x^{2}-2 x-5} d x=\int \frac{\frac{1}{2}(2 x-2)+4}{x^{2}-2 x-5} d x \\ & \quad=\frac{1}{2} \int \frac{2 x-2}{x^{2}-2 x-5} d x+4 \int \frac{1}{x^{2}-2 x-5} d x \end{aligned} $

Let $I_1=\int \frac{2 x-2}{x^{2}-2 x-5} d x$ and $I_2=\int \frac{1}{x^{2}-2 x-5} d x$

$\therefore \int \frac{x+3}{(x^{2}-2 x-5)} d x=\frac{1}{2} I_1+4 I_2$

Then, $I_1=\int \frac{2 x-2}{x^{2}-2 x-5} d x$

Let $x^{2}-2 x-5=t$

$\Rightarrow(2 x-2) d x=d t$

$\Rightarrow I_1=\int \frac{d t}{t}=\log |t|=\log |x^{2}-2 x-5|$

$I_2=\int \frac{1}{x^{2}-2 x-5} d x$

$=\int \frac{1}{(x^{2}-2 x+1)-6} d x$

$=\int \frac{1}{(x-1)^{2}+(\sqrt{6})^{2}} d x$

$=\frac{1}{2 \sqrt{6}} \log (\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}})$

Substituting (2) and (3) in (1), we obtain

$ \begin{aligned} \int \frac{x+3}{x^{2}-2 x-5} d x & =\frac{1}{2} \log |x^{2}-2 x-5|+\frac{4}{2 \sqrt{6}} \log |\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}|+C \\ & =\frac{1}{2} \log |x^{2}-2 x-5|+\frac{2}{\sqrt{6}} \log |\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}|+C \end{aligned} $

Solution

Let $5 x+3=A \frac{d}{d x}(x^{2}+4 x+10)+B$

$\Rightarrow 5 x+3=A(2 x+4)+B$

Equating the coefficients of $x$ and constant term, we obtain

$ \begin{aligned} & 2 A=5 \Rightarrow A=\frac{5}{2} \\ & 4 A+B=3 \Rightarrow B=-7 \\ & \therefore 5 x+3=\frac{5}{2}(2 x+4)-7 \\ & \Rightarrow \int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x=\int \frac{\frac{5}{2}(2 x+4)-7}{\sqrt{x^{2}+4 x+10}} d x \\ & \quad=\frac{5}{2} \int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x-7 \int \frac{1}{\sqrt{x^{2}+4 x+10}} d x \end{aligned} $

Let $I_1=\int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x$ and $I_2=\int \frac{1}{\sqrt{x^{2}+4 x+10}} d x$

$\therefore \int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x=\frac{5}{2} I_1-7 I_2$

Then, $I_1=\int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x$

Let $x^{2}+4 x+10=t$

$\therefore(2 x+4) d x=d t$

$\Rightarrow I_1=\int \frac{d t}{t}=2 \sqrt{t}=2 \sqrt{x^{2}+4 x+10}$

$I_2=\int \frac{1}{\sqrt{x^{2}+4 x+10}} d x$

$=\int \frac{1}{\sqrt{(x^{2}+4 x+4)+6}} d x$

$=\int \frac{1}{(x+2)^{2}+(\sqrt{6})^{2}} d x$

$=\log |(x+2) \sqrt{x^{2}+4 x+10}|$

Using equations (2) and (3) in (1), we obtain

$ \begin{aligned} \int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x & =\frac{5}{2}[2 \sqrt{x^{2}+4 x+10}]-7 \log |(x+2)+\sqrt{x^{2}+4 x+10}|+C \\ & =5 \sqrt{x^{2}+4 x+10}-7 \log (x+2)+\sqrt{x^{2}+4 x+10} \mid+C \end{aligned} $

Solution

$ \begin{aligned} \int \frac{d x}{x^{2}+2 x+2} & =\int \frac{d x}{(x^{2}+2 x+1)+1} \\ & =\int \frac{1}{(x+1)^{2}+(1)^{2}} d x \\ & =[\tan ^{-1}(x+1)]+C \end{aligned} $

Hence, the correct Answer is B.

Solution

$\int \frac{d x}{\sqrt{9 x-4 x^{2}}}$

$=\int \frac{1}{\sqrt{-4(x^{2}-\frac{9}{4} x)}} d x$

$=\int \frac{1}{-4(x^{2}-\frac{9}{4} x+\frac{81}{64}-\frac{81}{64})} d x$

$=\int \frac{1}{\sqrt{-4[(x-\frac{9}{8})^{2}-(\frac{9}{8})^{2}]}} d x$

$=\frac{1}{2} \int \frac{1}{\sqrt{(\frac{9}{8})^{2}-(x-\frac{9}{8})^{2}}} d x$

$=\frac{1}{2}[\sin ^{-1}(\frac{x-\frac{9}{8}}{\frac{9}{8}})]+C$

$(\int \frac{d y}{\sqrt{a^{2}-y^{2}}}=\sin ^{-1} \frac{y}{a}+C)$

$=\frac{1}{2} \sin ^{-1}(\frac{8 x-9}{9})+C$

Hence, the correct Answer is B.